Question: Let $y=\sec(x)$. What is the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi}{6}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2\sqrt3}{3}$ (Choice B) B $-\dfrac32$ (Choice C) C $-\dfrac{\sqrt{3}}{2}$ (Choice D) D $\dfrac23$
Explanation: Let's first find $\dfrac{dy}{dx}$. Then, we can evaluate it at $x=\dfrac{\pi}{6}$. Recall that the derivative of $\sec(x)$ is $\dfrac{\sin(x)}{\cos^2(x)}$, or $\sec(x)\tan(x)$. Put another way, $\dfrac{d}{dx}[\sec(x)]=\dfrac{\sin(x)}{\cos^2(x)}=\sec(x)\tan(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\dfrac{\pi}{6}}$ : $\begin{aligned} &\phantom{=}\dfrac{\sin\left({\dfrac{\pi}{6}}\right)}{\cos^2\left({\dfrac{\pi}{6}}\right)} \\\\ &=\dfrac{\dfrac12}{\left(\dfrac{\sqrt{3}}{2}\right)^2} \\\\ &={\dfrac12}\cdot\dfrac{4}{3} \\\\ &=\dfrac23 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi}{6}$ is $\dfrac23$.